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Chapter 4: Problem 70

Find the area of the region under the graph off on \([a, b]\). $$ f(x)=e^{-x / 2} ; \quad[-1,2] $$

Short Answer

Expert verified
The area of the region under the graph of the function \(f(x) = e^{-x/2}\) on the interval \([-1, 2]\) is approximately \(2(e^{1/2} - e^{-1})\).

Step by step solution

01

Identify the function and interval

We are given the function \(f(x) = e^{-x/2}\) and the interval \([-1, 2]\).
02

Integrate the function

We need to find the antiderivative of the given function, i.e., integrate: \[\int e^{-x/2} \, dx\] To make things easier, let's substitute \(u = -x/2\), so \(x = -2u\). Then, we have \(dx = -2 \, du\). Now, the integral becomes: \[\int e^u (-2) \, du\] We can now integrate with respect to \(u\): \[-2 \int e^u \, du\] We integrate \(e^u\) which will give us the antiderivative: \[-2(e^u + C) = -2(e^{-x/2} + C)\]
03

Calculate the definite integral

Now we need to find the definite integral over the interval \([-1, 2]\). This means we have to evaluate the antiderivative at the interval's endpoints and subtract them accordingly: \[ \begin{aligned} \int_{-1}^{2} e^{-x/2}\, dx &= -2\bigg|_{-1}^{2} (e^{-x/2}) \\ &= -2(e^{-2/2} - e^{1/2}) \\ &= -2(e^{-1} - e^{1/2}) \\ \end{aligned} \]
04

Simplify the expression

Simplifying the expression, we'll get the area of the region under the graph of the function: \[ \begin{aligned} \text{Area} &= -2(e^{-1} - e^{1/2}) \\ &= 2(e^{1/2} - e^{-1}) \end{aligned} \] So the area of the region under the graph of the function \(f(x) = e^{-x/2}\) on the interval \([-1, 2]\) is approximately \(2(e^{1/2} - e^{-1})\).

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